# Idempotent Ring is Commutative

## Theorem

Let $\struct {R, +, \circ}$ be an idempotent ring.

Denote with $0_R$ the zero of $R$.

Then $\struct {R, +, \circ}$ is a commutative ring.

## Proof

Let $x, y \in R$.

Then:

 $\displaystyle x + y$ $=$ $\displaystyle \paren {x + y}^2$ Definition of Idempotent Ring $\displaystyle$ $=$ $\displaystyle x^2 + x \circ y + y \circ x + y^2$ Binomial Theorem: Ring Theory $\displaystyle$ $=$ $\displaystyle x + x \circ y + y \circ x + y$ Definition of Idempotent Ring

Subtracting $x + y$ from both sides yields:

 $\displaystyle x \circ y + y \circ x$ $=$ $\displaystyle 0_R$ $\displaystyle$ $=$ $\displaystyle y \circ x + y \circ x$ Idempotent Ring has Characteristic Two

Finally, subtracting $y \circ x$ from both sides, we obtain:

$x \circ y = y \circ x$

and conclude $R$ is a commutative ring.

$\blacksquare$